KVPY Scholarship Test Exam Pattern

KVPY Scholarship Test Exam Pattern

The Kishore Vaigyanik Protsahan Yojana (KVPY) is a National Fellowship Program in Basic Sciences. It is open for undergraduate students, pursuing Graduation or Post Graduation studies in the field of Science. Students are selected from those who are studying in their XI standard to 1st year of any undergraduate Program in Basic Sciences. 

Also, students pursuing M.Sc./M.S. in Chemistry, Physics, Biology and Mathematics having an interest in scientific research are eligible for appearing in the KVPY Scholarship exam. 

Important Dates for the KVPY Scholarship Examination:

The notification for the KVPY Fellowship appears in all the national dailies normally on Technology Day (May 11) and the Second Sunday of July every year.

Selection Criteria for the KVPY Scholarship Examination 

Applicants will be selected on the following criteria: 

1. Selection of applicants based on the application form.
2. The selected students will then have to appear for the Aptitude Test which will be conducted both in Hindi and English languages. 
3. The short-listed students will be called for a personal interview, which is the final stage.

Important Note: The KVPY Scholarship will be provided based on both Aptitude tests and Interview marks.

KVPY Scholarship Test Exam Pattern – KVPY Scholarship SA Exam Pattern

For the stream SA, Part 1 consists of a total of 60 questions of 1 mark each. Part 2 consists of 20 questions of 2 marks each. The complete SA exam is of 100 marks.  

KVPY Scholarship SB/SX Exam Pattern

For the stream SB/SX, Part 1 consists of a total of 80 questions of 1 mark each. In this, the students can attempt any 3 sections. Part 2 consists of a total of 40 questions of 2 marks each. In this, the students can attempt any 2 sections. The complete SB/SX Exam is of 160 marks. 

The following table will help you understand the KVPY Scholarship Exam Pattern more deeply: 

KVPY Scholarship – Old Papers Exam Questions with Answers – KVPY Scholarship Test Exam Pattern

MATHEMATICS

Q1. Let ABC be an equilateral triangle with side length a. Let R and r denote the radii of the circumcircle and the incircle of triangle ABC respectively. Then, as a function of a, the ratio R/r

1. strictly increases
2. strictly decreases
3. remains constant
4. strictly increases for a < 1 and strictly decreases for a > 1

Solution: Let say a is the length of side of an equilateral triangle  

R – radii of the circumcircle of an equilateral triangle  

=> R =  a/√3

r – radii of the incircle of triangle

r = a/2√3

R/r = (a/√3)/(a/2√3)

=> R/r = 2

R/r = 2 is constant independent of a.

Hence, option C is correct.

Ans. (C) remains constant

Q2. Let N be the least positive integer such that whenever a no-zero digit c is written after the last digit of N, the resulting number is divisible by c. The sum of the digits of N is 

1. 9
2. 18
3. 27
4. 36

Solution: N is the least positive integer and when a digit c is written after the last digit of N, the resulting number is divisible by c.

⇒  So, N(c)/c

Least positive integer (N) which is divisible by digit c i.e. (1−9)

So possible number is

⇒  (9×8×7×5)c/c

Number, 9×8×7×5 is divisible by all the digits from 1 to 9

⇒  So, Number is 72×35=2520=N

⇒  Sum of the digits of number N=2+5+2+0=9

Ans. (A) 9

Q3. The number of cubic polynomials P(x) satisfying P(1) = 2, P(2) = 4, P(3) = 6, P(4) = 8 is

1. 0
2. 1
3. more than 1 but finitely many
4. infinitely many 

Solution: Let a Cubic polynomial P(x)=ax³+bx²+cx+d

Now,

P(1)=a+b+c+d=2                       —– ( 1 )

P(2)=8a+4b+2c+d=4                —– ( 2 )

P(3)=27a+9b+3c+d=6             —— ( 3 )

P(4)=64a+16b+4c+d=8           —— ( 4 )

Subtracting equation ( 1 ) from ( 2 ) we get,

⇒  7a+3b+c=2                —– ( 5 )

Subtracting equation ( 2 ) from ( 3 ) we get,

⇒  19a+5b+c=2             —– ( 6 )

Subtracting equation ( 5 ) from ( 6 ) we get,

⇒  12a+2b=0               —- ( 7 )

Subtracting equation ( 3 ) from ( 4 ) we get,

⇒  37a+7b+c=2              —— ( 8 )

Subtracting equation ( 6 ) from ( 8 ) we get,

⇒  18a+2b=0               —– ( 9 )

Subtracting equation ( 7 ) from ( 9 ) we get,

⇒  6a=0

∴   a=0

If a=0 then, the polynomial will be

P(x)=bx²+cx+d

Hence, there is no cubic polynomial possible.

Ans. (A) 0 

PHYSICS

Q1. Various optical processes are involved in the formation of a rainbow. Which of the following provides the correct order in time in which these processes occur?

1. Refraction, total internal reflection, refraction.
2. Total internal reflection, refraction, total internal reflection.
3. Total internal reflection, refraction, refraction.
4. Refraction, total internal reflection, total internal reflection.

Solution: Rainbow is formed due to total internal reflection of incident light by liquid drops.

Hence, the process is :

1. Refraction of incident ray
2. Total internal reflection
3. Again refraction when rays come out of liquid drops.

Ans. (A) Refraction, total internal reflection, refraction.

Q2. A specially designed Vernier calliper has the main scale least count of 1 mm. On the Vernier scale, there are 10 equal divisions, and they match with 11 main scale divisions. Then, the least count of the Vernier calliper is

1. 0.1 mm
2. 0.909 mm
3. 1.1 mm
4. 0.09 mm

Solution: Least count = 1 V.S.D.−1 M.S.D.

V.S.D → Vernier scale division

M.S.D → Main scale division

L.C = 11/10 – 1

      = 11-10/10 = 1/10 = 0.1 mm

Ans. (A) 0.1 mm

Q3. Frosted glass is widely used for translucent windows. The region where a transparent adhesive tape is stuck over the frosted glass becomes transparent. The most reasonable explanation for this is

1. The diffusion of adhesive glue into the glass.
2. The chemical reaction at the adhesive tape – glass interface.
3. The refractive index of adhesive glue is close to that of glass.
4. adhesive tape is more transparent than glass

Solution: Frosted glass has different refractive index on its surface as frosted glass is produced by coating on the surface. Now, as a transparent adhesive tape is stuck on the glass surface, the refractive index of the upper coating is adjusted to be the same as that of glass, so it becomes transparent now. So, the answer is (C).

Ans. (C) the refractive index of adhesive glue is close to that of glass.

CHEMISTRY

Q1. In water-gas shift reaction, hydrogen gas is produced from the reaction of steam with

1. methane
2. coke
3. carbon monoxide
4. carbon dioxide

Solution: Water-gas shift reaction explains the reaction of carbon monoxide and water vapour to form carbon dioxide and hydrogen:

CO+H₂O→CO₂+H₂

Hence, option C is correct.

Ans. (C) carbon monoxide

Q2. The element whose salts cannot be detected by the flame test is

1. Mg
2. Na
3. Cu
4. Sr

Solution: Mg cannot be detected by flame test due to high ionisation energy.

Ans. (A) Mg

Q3. During the free expansion of an ideal gas in an isolated chamber,

1. Internal energy remains constant
2. Internal energy decreases
3. Work done on the system is negative
4. Temperature increases

Solution: q = 0 

P_ext  = 0 ( for free expansion) => Work done is zero

Therefore, internal energy remains constant.

Ans. internal energy remains constant

BIOLOGY

Q1. Which one of the following is a CORRECT statement about primate evolution?

1. Chimpanzees and gorillas evolved from macaques
2. Humans and chimpanzees evolved from gorillas
3. Humans, chimpanzees and gorillas evolved from a common ancestor
4. Humans and gorillas evolved from chimpanzees

Ans. (C) Humans, chimpanzees and gorillas evolved from a common ancestor 

Q2. Removal of the pancreas impairs the breakdown of

1. lipids and carbohydrates only
2. lipids and proteins only
3. lipids, proteins and carbohydrates
4. proteins and carbohydrates only

Ans. (C) lipids, proteins and carbohydrates

Q3. Microscopic examination of a blood smear reveals an abnormal increase in the number of granular cells with multiple nuclear lobes. Which one of the following cell types has increased in number?

1. Lymphocytes
2. Monocytes
3. Neutrophils
4. Thrombocytes

Ans. (C) Neutrophils